Problem: What is the slope of the secant line that intersects the graph of $h(x)=\sqrt{15-2x}$ at $x=3$ and $x=7$ ?
The secant line will pass through points $(3,h(3))$ and $(7,h(7))$. $\begin{aligned} \text{slope}&=\dfrac{\text{change in }y}{\text{change in }x} \\\\ &=\dfrac{h(7)-h(3)}{7-3} \end{aligned}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${1}$ ${2}$ ${3}$ ${4}$ $y$ $x$ $(3,h(3))$ $(7,h(7))$ secant line We will need to know the values of $h(3)$ and $h(7)$ to find the slope. $\begin{aligned} h(3)&=\sqrt{15-2(3)} \\\\ &=3 \\\\ h(7)&=\sqrt{15-2(7)} \\\\ &=1 \end{aligned}$ Now we can find the slope: $\begin{aligned} \dfrac{h(7)-h(3)}{7-3}&=\dfrac{1-3}{4} \\\\ &=-\dfrac{1}{2} \end{aligned}$ The slope of the secant line that intersects the graph of $h(x)=\sqrt{15-2x}$ at $x=3$ and $x=7$ is $-\dfrac{1}{2}$. Notice that the slope of the secant line is calculated just like the average rate of change over the interval.